Calculating Displacement in Two-Dimensional Motion

  • Thread starter Thread starter Gamegoofs2
  • Start date Start date
  • Tags Tags
    Motion
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 2K views
Gamegoofs2
Messages
3
Reaction score
0
I'm having trouble with adding two-dimensional motion. Here's an example of the type of problem I'm having trouble with:

A map gives you the following directions:

Travel 45 mintues at a speed of 1.2 m/sec and a heading of 45 degrees. Turn to a heading of 330 degrees and travel at the same speed for 32 minutes.
What will your displacement be if you follow these directions?

I was taught to use this equation:
[tex]\Delta[/tex]x = initial velocity * time + 1/2 * acceleration * time^2
and
Ax= initial velocity * cos()
Ay= initial velocity * sin()
and the same for the the second vector.

I started out with trying to find the first vector or vector A as I called it.
[tex]\Delta[/tex]x= 1.2 m/sec(2700 sec) + 1/2 (0)(2700 sec)^2
and got [tex]\Delta[/tex]x=3240m which I thought was a bit high, so I don't know if I'm doing something wrong or if it should really be that high.
 
on Phys.org
Welcome to PF!

Hi Gamegoofs! Welcome to PF! :smile:

(have a delta: ∆ :wink:)
Gamegoofs2 said:
Travel 45 mintues at a speed of 1.2 m/sec and a heading of 45 degrees.

I started out with trying to find the first vector or vector A as I called it.
[tex]\Delta[/tex]x= 1.2 m/sec(2700 sec) + 1/2 (0)(2700 sec)^2
and got [tex]\Delta[/tex]x=3240m which I thought was a bit high, so I don't know if I'm doing something wrong or if it should really be that high.

No, that's ok …

and it's not high, it's only 3km after nearly an hour! :wink: